Question

i need help with part two

PART TWO 1) The steady-state kinetics of an enzyme is studied in the absence and presence of inhibitor A. The initial rate is given as a function of substrate concentration in the following table. [S] (MM) * Velocity in substrate only y (mM/min) .25 1.72 58 ,598 r.60 2.04 .44 50 A 2.63 238 5.00 .2 3.33 10.00 4.17 4 Velocity in substrate + S inhibitor A (MM/min) 0.98 1.02 | 1.17 . 5 | 1.47 . 1.96 sl 2.38 . (a) Draw the Lineweaver-Burk plot on the graph paper provided using the kinetic data above. Label all the axes on your graph. (5 pts) & On back attachment & (b) Using the above Lineweaver-Burk plot, determine Vmes and Km in the absence and presence of inhibitor A. Also, determine the type of inhibition involved? (5 pts) Non competitive inhibition (1) Page 4

Answer 1

a) The Lineweaver-Burk plot is the plot of 1/V₀ vs. 1/[S] so, you obtain the following graph:

And the linear regression equations are:

V [min] substrate tinhibitor A Pttttt substrate only 0.2 0.4 0.6 0.8 *iš [mi] -0.4

Substrate only:

$y=0.4843x+0.1951$

Substrate plus inhibitor A:

$y=0.8582x+0.3369$

Where the y-intercept corresponds to the inverse of the maximum velocity and the slope to the ratio K_m/V_max so, for the substrate only we have:

$y-intercept=\frac{1}{V_{max}}\Rightarrow V_{max}=\frac{1}{y-intercept}=V_{max}=\frac{1}{0.1951min/mM}=5.13mM/min$

$slope=\frac{K_m}{V_{max}}\Rightarrow K_m=slope*V_{max}=0.4843min*5.13mM/min=2.48mM$

For the substrate plus inhibitor we have:

$y-intercept=\frac{1}{V_{max}}\Rightarrow V_{max}=\frac{1}{y-intercept}=V_{max}=\frac{1}{0.3369min/mM}=2.97mM/min$

$slope=\frac{K_m}{V_{max}}\Rightarrow K_m=slope*V_{max}=0.8582min*2.97mM/min=2.55mM$

As both graphs have approximately the same value of K_m , we have a noncompetitive inhibition, this can also be seen in the graph: both lines intercept the x-axis at the same point, ~ -0.40 mM^-1 , the negative inverse of K_m:

$x-intercept=-\frac{1}{K_m}=-\frac{1}{2.48}=-0.39mM^{-1}$

$x-intercept=-\frac{1}{K_m}=-\frac{1}{2.55}=-0.40mM^{-1}$

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